Assuming the variances are equal, use the HSB data set to compare the means of m

Question

Assuming the variances are equal, use the HSB data set to compare the means of male vs. female students on the (a) the science achievement score, and (b) the writing achievement score. Again, test the null hypothesis vs. the alternate hypothesis for each respective achievement score. Conduct this tests at = 0.05. For each achievement score report the values of Y¯ 1, Y¯ 2, the standard errors of the two means, sY¯1 and sY¯2 , the sample sizes n1, n2, the value of the test statistic, t, and the pvalue associated with this value of t. Clearly state your decision regarding the null hypothesis and your conclusion in the context of the problem. Write down the 95% confidence interval for µ1 µ2 and interpret this interval.

Group Statistics Std. DeviationStd. Error Mean 10.085 9.217 10.176 SCI Male 273 53.23 610 WRTG Male 273 49.79 616 327 54.55 Independent Samples Test Levene's Test for Equality of arian f Mean Sig. 95% Confidence Interval (2 Mean Std. Error df tailed) Difference Difference Lower 2.691 SCI Equal variances assumed 1.707 192 3.412 598 001 789 1.142 4.240 Equal variances not 3.384 557.447001 2.691 795 1.129 4.253 WRTG Equal variances assumed 12.682 .000 59B 0004.768 774 -3.249 6.162 6.288 Equal variances not 540.781000 .768 .784 -3.22B assumed 6.081 6.309

Explanation / Answer

For the science achievement score

H0: 1 - 2 = 0 i.e. (1 = 2)

H1: 1 - 2 0 i.e. (1 2)

            

Assuming population variances are equal, we would have to calculate pooled-variance t-Test

Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)

         = (273-1)*10.085^2+(327-1)*9.217^2/272+326

         = 92.57

tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)

       =(53.23-50.54)-0/92.57(1/273+1/327)

       =2.69/0.7888

       =3.41

tCRIT is +/- 1.96 and hence reject the null hypothesis

(X1-X2)+-ta/2*Sp^2(1/n1+1/n2)

2.69+/-1.96*0.7888

LCL=2.69-1.546048=1.1439

UCL=2.69+1.546048=4.236

(1.1439,4.236)

For the writing achievement score

H0: 1 - 2 = 0 i.e. (1 = 2)

H1: 1 - 2 0 i.e. (1 2)

            

Assuming population variances are equal, we would have to calculate pooled-variance t-Test

Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)

         = (273-1)*10.176^2+(327-1)*8.778^2/272+326

         = 89.1057

tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)

       =(49.79-54.55)-0/89.1057(1/273+1/327)

       =-6.15

tCRIT is +/- 1.96 and hence reject the null hypothesis

(X1-X2)+-ta/2*Sp^2(1/n1+1/n2)

-4.76+/-1.96*0.77387919

LCL=-4.76-1.5168=-6.278

UCL=-4.76+1.5168=-3.2432

(-6.278, -3.2432)

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